The work done by the man against friction is 4,192.86 J.
The given parameters;
force applied, F = 540 Nangle of inclination, θ = 75⁰horizontal distance, x = 30 mThe work done by the man against friction is calculated as follows;
[tex]W = F \times d \times cos(75)\\\\W = 540 \times 30 \times cos(75)\\\\W = 4,192.86 \ J[/tex]
Thus, the work done by the man against friction is 4,192.86 J.
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Accelerations are produced by
A. Masses
B.accelerations
C. Velocities
D.unbalanced, net forces
difference between speed and velocity
Answer:
Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement.
Explanation:
A gold doubloon 6.1 cm in diameter and 2.0 mm thick is dropped over the side of a pirate sheep. When it comes to rest on the ocean floor at a depth of 770 m, how much has its volume changed?
The definition of volume modulus and the variation of pressure with depth allows to find the result for the variation of the volume of the coin is:
ΔV = 2.15 10⁻⁸ m³
The pressure with the depth is given by the relation
P = P₀ + ρ g h
Where P is the pressure, ρ is the density anf h depth.
The size of the bodies is determined by the distance of their atomic and molecular bonds, therefore the size of the bodies changes under external interations, in the case of hydrostatic pressure a constant called volumetric modulus is defined.
[tex]B = - \frac{\Delta P}{\frac{\Delta V}{Vo} } \\\Delta V = - \frac{\Delta P }{B} \ V_o[/tex]
Where ΔP is the pressure change, V₀ and V are the volume change and the initial volume of the body, the negative sign is introduced so that the volumetric modulus is a positive quantity.
They indicate the diameter and thickness of the coin (d = 6.1 cm and e =0.20 cm) on the sea surface and the depth to which it is submerged
h = 770 m
Let's look for the volume of the coin.
V₀ = π r² h = [tex]\pi \ \frac{d^2}{4} \ e[/tex]
V₀ = [tex]\pi \ \frac{0.061^2 }{4} \ 0.002[/tex]
V₀ = 5.84 10-6 m³
Let's find the pressure at the depth of y = 770 m, the density of sea water is ρ = 1025 kg / m³, the pressure at the surface is the atmospheric pressure P₀ = 1 10⁵ Pa, the volumetric modulus of water is B = 0.21 10¹⁰ Pa.
P = 1 10⁵ + 1025 9.8 770
P = 1 10⁵ + 7,735 10⁶
P = 7.84 10⁶ Pa
Let's calculate
ΔV =[tex]- \frac{1 \ 10^5 - 7.84 \ 10^6 }{0.21 \ 10^{10}} \ 5.845 \ 10^{-6}[/tex]
ΔV = 2.15 10-8 m³
In conclusion using the definition of volume modulus and the variation of pressure with depth we can find the result for the variation of the volume of the coin is:
ΔV = 2.15 10-8 m³
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calculate the surface area of a box whose mass is 200 kg and exerts a pressure of 100 Pascal on the floor.
Answer:
Explanation:
If 2×2 is 4 so 1 kg can be 1 gram if it belive on it self some people change
The linear distance traveled by a wheel of radius 50cm after 99 complete revolutions is?
1)99m
2)210m
3)311
4)433
Answer:
3) 311 m
Explanation:
Circumference = 2πR = π m/rev
99 rev(π m/rev) = 99π m or about 311 meters
Carbon tetrachloride (CCl4) is diffusing through benzene (C6H6), as the drawing illustrates. The concentration of CCl4 at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s. Using these data and those shown in the drawing, find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A.
We have that for the Question "find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A."
Answers:
Mass of CCI_4 per second = [tex]5.86*10^{-13} kg/s[/tex] Concentration of CCI_4 = [tex]12.6*10^{-3}kg/m^3[/tex]
From the question we are told
The concentration of [tex]CCl_4[/tex] at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s
A) the mass flow rate of CCI_4 as it passes point A is the same as the mass flow rate at which CCI_4 enters the left end of the tube
Therefore, the mass flow rate of CCI_4 at point A
= [tex]5.86*10^{-13} kg/s[/tex]
B) From Fick's law
[tex]\deltaC = \frac{mL}{DAt}\\\\ Assume L = 5*10^{-3}, A = 3*10^{-4}\\\\\deltaC = \frac{5.86*10^{-13} * 5*10^{-3}}{21.9*10^{-10} * 3*10^{-4}}\\\\\deltaC = 4.46*10^{-3}kg/m^3[/tex]
Then,
[tex]Concentration = 1.71*10^{-2} - 4.46*10^{-3}\\\\= 12.6*10^{-3}kg/m^3[/tex]
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Read the scenario below and answer the question that follows. Randall is hiring cooks for his restaurant. The first applicant is a handsome man with an average resumé and average job experience. The second applicant is a far less attractive man with a slightly above average resumé and above average job experience. Randall decides to hire the first applicant. Based on this information and on Randall’s decision, what might a psychologist conclude about Randall’s social perception? Randall has an unconscious assumption that attractive people are more competent. Randall has a unconscious assumption that unattractive people are bad cooks. Randall has a conscious assumption that attractive people make better cooks. Randall has a conscious assumption that unattractive people are more competent.
Randall has an unconscious assumption that attractive people are more competent.
What is meant by assumption ?The term assumption can be described as an unspoken premise that underlies the conclusion.
Here,
The capacity to accurately evaluate and draw conclusions about other individuals based on their overall physical appearance, verbal behaviour, and nonverbal attitudes is referred to as social perception.
Given that, for his restaurant, Randall is employing chefs. The first applicant is a dashing man with an average resume and career history. The second candidate is a far less appealing man with an average to slightly above average resume and work experience. Randall chooses to hire the first candidate.
This shows the social perception of Randall and his unconscious assumptions against unattractive people.
Hence,
Randall has an unconscious assumption that attractive people are more competent.
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help :”)
a skydiver jumps out of a plane and falls for 45 seconds before deploying his parachute. how far did he fall?
Answer:
200 feet
Explanation:
A small, free-to-rotate magnet is placed in a strong magnetic field. In what orientation will it come to rest
Answer:
South-North
Explanation:
73 ml of water is followed by 25 ml of juice. What is the percent strength of juice?
Total liquid: 73 + 25 = 98 ml
Percent juice = (25/98) x 100 = 25.5 %
3) A 60. kg person is in an elevator. The elevator starts from rest and then accelerates upwards at 2.0 m/s^2 for 4.0 seconds. Calculate the work done by the normal force on the person. *
Answer:
WD = 960 J
Explanation:
WD = work done (J)
F = force (N)
s = displacement (m)
m = mass (kg) = 60
a = acceleration (m/s²) = 2
t = time (s) = 4
u = initial velocity (m/s) = 0
The formulas or equations that are relevant ate:
WD = F × s
F = m × a
s = u + at
We want to find WD, so we need to now the force and the displacement (or distance);
We calculate force, in Newtons, with the formula F = ma:
F = 60 × 2
F = 120 N
We also need displacement, which get with the formula s = u + at:
s = 0 + 2(4)
s = 8 m
Now we have F and s, we can calculate WD:
WD = 120 × 8
WD = 960 J
Methodology:
Starting with what you want to find, in this case WD, list the formula/s you could use;
Then, identify the information you need for the formula and whether or not you are given that information;
Next, list the formulas for the information you don't have and once again, identify whether the information you are given is sufficient to use those formulas;
Once you can calculate all necessary information, then proceed to calculate the values and finally, the answer;
I suggest also keeping a list of all the variables as I've done at the top of my working so it is clear for you to see and use.
How does friction help us in walking.
science thanks sa points
Answer: Are these free point?
Explanation:
A race car traveling at 100 m/s enters an unbanked turn of 400 m radius. The coefficient of (static) friction between the tires and the track is 1.1. The track has both an inner and an outer wall. Which statement is correct
Answer:
The race car will crash into the outer wall
Explanation:
max fr = μsN = 1.1 mg = 11 m
mv2/R = m(100)2/(400) = 25 m > fr
What is discordant characteristic ?
[tex] \: \: \: \: [/tex]
being at variance; disagreeingor incongruous: discordant opinions. disagreeable to the ear; dissonant; harsh.hope it helps[tex] \: \: \: \: \: [/tex]
dissimilar with respect to one or more particular characters
The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.2 m above the ground, how steep a slope can the truck be parked on without tipping over
The slope of the road can be given as the ratio of the change in vertical
distance per unit change in horizontal distance.
The maximum steepness of the slope where the truck can be parked without tipping over is approximately 54.55 %.Reasons:
Width of the truck = 2.4 meters
Height of the truck = 4.0 meters
Height of the center of gravity = 2.2 meters
Required:
The allowable steepness of the slope the truck can be parked without tipping over.
Solution:
Let, C represent the Center of Gravity, CG
At the tipping point, the angle of elevation of the slope = θ
Where;
[tex]tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}[/tex]
The steepness of the slope is therefore;
[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100[/tex]
Where;
[tex]\overline{AM}[/tex] = Half the width of the truck = [tex]\dfrac{2.4 \, m}{2}[/tex] = 1.2 m
[tex]\overline{CM}[/tex] = The elevation of the center of gravity above the ground = 2.2 m
[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%[/tex]
[tex]tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}[/tex]
[tex]Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}[/tex]
The maximum steepness of the slope where the truck can be parked is 54.55 %.
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If the voltage across a 5-F capacitor is 2*e^-3
V find the current and the power
What is the relationship between balancing equations and the law of conservation of matter
Which is the main gas that makes up the Earth's atmosphere?
Answer:
78 percent nitrogenExplanation:
I hope it's helpful for you
A man is whirling a 0.25 kg ball on a 1.5 m long string at 3 m/s. Find the centripetal acceleration of this ball.
Question 2 options:
0.5 m/s2
13.5 m/s2
6 m/s2
2 m/s2
The centripetal acceleration of this ball is equal to 12 [tex]m/s^2[/tex]
Given the following data:
Diameter = 1.5 mSpeed, V = 3 m/s.Mass = 0.25 kgRadius = [tex]\frac{Diameter}{2} = \frac{1.5}{2} = 0.75 \;meters[/tex]
To find the centripetal acceleration of this ball:
The acceleration of an object along a circular track is referred to as centripetal acceleration.
Mathematically, the centripetal acceleration of an object is given by the formula:
[tex]A_c = \frac{V^2}{r}[/tex]
Where:
Ac is the centripetal acceleration.r is the radius of the circular track.V is the velocity of an object.
Substituting the given parameters into the formula, we have;
[tex]A_c = \frac{3^2}{0.75}\\\\A_c = \frac{9}{0.75}\\\\A_c = \frac{9}{0.75}[/tex]
Centripetal acceleration = 12 [tex]m/s^2[/tex]
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a student lifts a toy car from a bench and places the toy car at the top of a slope describe an energy transfer that occurs when the student lifts the toy car from the bench and places the toy car at the top of the slope.
Answer:
Assuming there are no energy losses due to friction or drag, the gravitational potential energy will change into kinetic potential energy as the car reaches the bottom of the slope.
G.P.E = m*g*h
K.E = (m*v^2)/2
where
m = mass of toy car (kg)
g = gravity (m/s^2)
h = heigh of your car from the bottom (m)
v = velocity of the toy car as it reaches the bottom (m/s)
Equate K.E to G.P.E
G.P.E = K.E
m*g*h = (m*v^2)/2
make v the subject of the formula
v = (2*g*h)^(1/2)
Substitute g = 9.81 m/s^2 and h = 2m into the equation to get v
v = (2*9.81*2)^(1/2)
v = 6.264 m/s
Suppose a ball is thrown vertically upward (positive direction) from an initial height LaTeX: h_0 with initial velocity LaTeX: v_0. Find the position function LaTeX: s(t) of the ball after LaTeX: t seconds assuming the gravitational acceleration LaTeX: g is a positive constant pointing downward (negative direction).
After time t, the position function of the ball is determined as [tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]
The given parameters;
initial velocity of the ball, = [tex]v_0[/tex]initial position of the ball, = [tex]h_0[/tex]acceleration due to gravity, = gThe position function of the ball after time t, is calculated as follows;
[tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]
The negative sign of acceleration of due to gravity is because the ball is moving upward against gravity.
Thus, after time t, the position function of the ball is determined as [tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]
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Does it appear that true average HAZ depth is larger for the high current condition than for the nonhigh current condition
Answer:
The data suggest that the true mean HAZ depth is larger when the current setting is higher.
is a fuel cell a primary or secondary cell
Answer:
Enrol in our 50 studyscore masterclass. Click here! Fuel cells are a type of primary cell in that they are not recharged, However, they are unique because they never run out, if the reactants are constantly supplied.
An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time?
Free-fall Acceleration is -10 m/s^2
Answer:
Explanation:
s = s₀ + v₀t + ½at²
s = 0 + 0(15) + ½(6)(15²)
s = 675 m
Not sure what the free fall acceleration is needed for, but if the object is dropped from a high enough point, it will travel in 15 seconds
s = ½10(15²) = 2250 m if air resistance is ignored
Some students conduct an experiment to prove conservation of momentum. They use two objects that collide Measurements
are taken before and after the collision.
Which two quantities will the students multiply together before and after the collision?
A. mass and velocity
B. distance and time
C. mass and acceleration
D. velocity and time
This question involves the concepts of the law of conservation of momentum, velocity, and mass.
The two quantities, the students should multiply before and after the collision are "A. mass and velocity".
According to the law of conservation of momentum, In an isolated system, the total momentum of the system before the collision is always equal to the total momentum of the system after the collision.
To prove the law of conservation of momentum, consider two balls of masses ‘m₁’ and ‘m₂’, moving with velocities ‘u₁’ and ‘u₂’, respectively, such that u₁ is greater than u₂. After some time, these balls collide with each other and their velocities become ‘v₁’ and ‘v₂’, respectively.
This situation is illustrated in the attached picture.
So, according to the law of conservation of momentum:
Total Momentum Before Collision = Total Momentum After Collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
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Sorry this is a year late, but here it is for those of you who are stuck on the same thing.
======================================Proving Conservation of Momentum Quick Check - 5/5NOTE: Please Check and Confirm That You Are On The Same Assignment with The Same Questions and Number of Questions. Thank You and Good Luck!
=======================================1. Mass & velocity
2. The total momentum after the collision is the same as the total momentum before the collision.
3. 0.54 kg⋅m/s
4. The system has external forces, such as friction and air resistance, acting on it.
5. 3.0 m/s
To get the dimmest bulbs with two batteries and two bulbs you would connect the batteries in ____ and the bulbs in ____.
Answer:
batteries in parallel connection and bulbs in serial connection
To get the dimmest bulb with two batteries and two bulbs you would connect the batteries in parallel and the bulbs in series.
What is Parallel and series circuits?When two-terminal components and electrical networks that can be connected in series or parallel. This will result in two terminals in the electrical network, and may themselves participate in a series or parallel topology. When a two-terminal "object" is an electrical component or electrical network is a matter of perspective.
A circuit is said to be in series when the same current flows through all the components in the circuit where the current has only one path. A circuit is said to be parallel when there are multiple paths for the electric current to flow through it where the components which are part of the parallel circuit will have a constant voltage across all their ends.
Thus, to get the dimmest bulb with two batteries and two bulbs you would connect the batteries in parallel and the bulbs in series.
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the c component of vector a is 5.3 units, and it’s y component is -2.3 units. the angle that vector a makes with the +x axis is closest to
110
160
23
340
250
Answer:
340
Explanation:
Sorry I don't know how to do this one yet, I just found the answer in a textbook.
The angle that vector a makes with the +x axis is closest to 23.
What is direction of a vector?The direction of a vector is represented tangent of angle equal to the ratio of the y component and the x component of the vector quantity.
tangent of angle = y/x
angle = tan⁻¹ (-2.3/5.3)
angle = 23.46°
Thus, the angle that vector makes with +x is 23.
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Describe the concept of energy quanta of EM radiation which was explained by Planck.
Answer:
Planck postulated that the energy of light is proportional to the frequency, and the constant that relates them is known as Planck's constant (h). His work led to Albert Einstein determining that light exists in discrete quanta of energy, or photons.
Explanation:
Answer:
Energy does not occur in continuous amounts but in discrete amounts described by:
E = N h ∨ where N is the number of quanta (energy units), ∨ the frequency of the energy, and h Planck's constant (6.63E-34 J-sec)
please help me
please help me
please help me
Answer:
do it got a picture
on the edge
Explanation: